harmonic forcing under natural frequency 1

[hechengli]

for a second order differential equation,( mass-spring-damping) system, if we have harmonic force acting on the system, we can find the solution x(t), but how about if the frequency of harmonic force is the natural frequency? there will be resonant. we know it won't go to infinity because of damping. Are there any formula availabe for the response x(t) of the system under this kind of force? Thanks!

 

[electricpete]

Here's what I think:

at resonance the mass accereration and spring force terms cancel.

We are left with F=C d/dt(X)
Assume X has the form X0sin(w0*t) where w0=sqrt(k/m)

F=C sqrt(K/M)X0 cos (w0t)

The magnitude of peak displacement X0 will be
X0=F/[C*sqrt(K/M)] (where F is peak force).

The phase of the displacement will be lagging F by 90 degrees.

 

[GregLocock]

That's the fastest derivation of that I've seen, there is a slight niggle in that it doesn't quite predict the maximum amplitude, as your assumption for the resonant frequency (an admittedly nebulous term) is for the undamped system. The answer you get is close enough for all practical purposes, I suspect.

Cheers

Greg Locock

 

[hechengli]

thanks for the reply!
what I want is the analytical solution to the harmonic forcing at the natural frequency, find x(t)=F(t,c,m,k), which can be drawn against the time t,and the amplitude of x will go from zero to a certain value, state state. is that easy, I haven't find any book have the discussion on this topic.

 

[electricpete]

I agree with Greg's comment that the solution is approximate but will be pretty close for lightly-damped systems (probably as close as our limited ability to estimate the actual damping of a real-world system warrants).

I think we have given you what you asked for. The functional form of the solution is a sinusoid. The magnitude and phase have been discussed.

If I get a chance I'll try to work on the more precise solution... which won't change the functional form.

 

[electricpete]

Here is an attempt at the more exact solution... with a little help from my computer (Maple). Hopefully it comes out legible.

Differential equation:

f(t)=m*diff(x(t),t,t)+c*diff(x(t),t)+m*x(t);
/ 2 |d | /d f(t) = m |--- x(t)| + c |-- x(t)| + m x(t)
| 2 | \dt /
\dt /

Transfer function H(w)=X(w)/F(w):
> H(w):=m*(I*w)^2+c*(I*w)+k;

2
H(w) := -m w + I w c + k

Magnitude of transfer function:
> Hmag:=sqrt((k-m*w^2)^2+(w*c)^2);

2 2 4 2 2 2 1/2
Hmag := (k - 2 k m w + w m + w c )

Derivative of magnitude of transfer function wrt w
> Hmagdiff:=diff(Hmag,w);

3 2 2
-4 k m w + 4 w m + 2 w c
Hmagdiff := 1/2 ----------------------------------
2 2 4 2 2 2 1/2
(k - 2 k m w + w m + w c )

Find the resonance at point where Hmag is max (deriv=0):
> wr:=solve(Hmagdiff=0,w);

2 1/2
(4 k m - 2 c )
wr := 1/2 -----------------
m

> Hmagresonance:=subs(w=wr,Hmag);

/ 2 2 2
| 2 k (4 k m - 2 c ) (4 k m - 2 c )
Hmagresonance := |k - 1/2 ---------------- + 1/16 ---------------
| m 2
\ m

2 2\1/2
(4 k m - 2 c ) c |
+ 1/4 -----------------|
2 |
m /

> Hangle_resonance:=arctan(Im(H(w))/Re(H(w)));

2
Im(m w - I w c - k)
Hangle_resonance := arctan(--------------------)
2
Re(m w - I w c - k)
The solution X will have a magnitude given by the magnitude of H at resonance times the magnitude of F. The angle of X with respect to F is given by Hangle_resonance above.

 

[GregLocock]

Ah, I think hechengli perhaps wants to see the transient response as the harmonic force signal is applied? Obviously in the steady state it will tend towards your answer.

That is, he may be interested in the response to the excitation F(t)=if(t<start,0,sin(w*t)), where start>0 and w=sqrt(k/m)

if so that is a lot more interesting. So hechengli - do you want the transient response or the steady state response, which pete's already done for you? incidentally, are we doing your homework for you? Cheers

Greg Locock

 

[electricpete]

Those computer-equations didn't work. Let me use the other format:
Differential equation:

f(t)=m*diff(x(t),t,t)+c*diff(x(t),t)+m*x(t);f(t) = m*diff(diff(x(t),t),t)+c*diff(x(t),t)+m*x(t)

Transfer function H(w)=X(w)/F(w):
> H(w):=m*(I*w)^2+c*(I*w)+k;
H(w) := -m*w^2+I*w*c+k

Magnitude of transfer function:
> Hmag:=sqrt((k-m*w^2)^2+(w*c)^2);
Hmag := (k^2-2*k*m*w^2+w^4*m^2+w^2*c^2)^(1/2)

Derivative of magnitude of transfer function wrt w
> Hmagdiff:=diff(Hmag,w);
Hmagdiff := 1/2/(k^2-2*k*m*w^2+w^4*m^2+w^2*c^2)^(1/2)*(-4*k*m*w+4*w^3*m^2+2*w*c^2)

Find the resonance at point where Hmag is max (deriv=0):
> wr:=solve(Hmagdiff=0,w);
wr := 1/2*(4*k*m-2*c^2)^(1/2)/m

> Hmagresonance:=subs(w=wr,Hmag);
Hmagresonance := (k^2-1/2*k/m*(4*k*m-2*c^2)+1/16*(4*k*m-2*c^2)^2/m^2+1/4*(4*k*m-2*c^2)/m^2*c^2)^(1/2)


> Hangle_resonance:=arctan(Im(H(w))/Re(H(w)));
Hangle_resonance := arctan(-w*c/(m*w^2-k) (I think)

The solution X will have a magnitude given by the magnitude of H at resonance times the magnitude of F. The angle of X with respect to F is given by Hangle_resonance above

 

[electricpete]

wr has the pleasing property that it reduces to sqrt(k/m) when c=0.

 

[electricpete]

.... but the unpleasant property that with c=0 Hangle=arctan(0)=0.... should be pi/2. You'll have to find the error yourself.

 

[hechengli]

thanks again!
what I want to see is the transient response of the system under the harmonic force at its natural frequency. Not the steady state. When I do experiment, I found the magnitude goes from zero to steady state, so I am kind of interested in find the exact solution for a damped system, maybe transient response as Greg Locock said.

 

[GregLocock]

This problem has a very ugly analytical solution, the step function in the forcing makes it a tricky case. I have a strong suspicion that a Laplace transform will be the easiest analytical approach. Note that this forcing function is not really a harmonic excitation, you get a fair bit of high frequency stuff at the switch-on.

My Excel simulation in the time domain looks plausible, but seems to be giving the wrong answers. grr

Cheers

Greg Locock

 

[electricpete]

Ok, I found there error in my steady state solution. I typed in H(w):=m*(I*w)^2+c*(I*w)+k... when that is actually the inverse of H(w). Fortunately I still found the correct wr when I set d/dw=0 I found the min of the inverse which is still the max of the transfer function. All that remains for the steady state solution is to substitute my value of wr into the correct H(w)=1/[m*(I*w)^2+c*(I*w)+k].

The transient solution as Greg said would probably best be done by Laplace. If I get a chance I'll let my computer try. It would of course be easier if you have in mind specific values of m, c, k and the phase theta of the force F=F0cos(wrt+theta) at the moment it is &quot;turned on&quot;.

 

[pja]

Every vibrations book ever published has the closed form solution to this problem derived,rederived and rerederived 8000 times over..go to your nearest techicial library and check it out.

 

[GregLocock]

Perhaps it does (tho I don't remember seeing the transient behaviour in an analytical form). But this is fun, for a strange value of fun.
Cheers

Greg Locock

 

[electricpete]

Maple tackled the transient problem pretty easily with laplace transforms. The solution as Greg said is somewhat ugly, but when plotted with sample values the parameters it looks reasonable. It's somewhat more cumbersome than what I posted above. I'll post it in same format if you want. In meantime I'll try to figure out a way to create a word file that captures the equations without mangling them.

 

[electricpete]

I've spent some time playing with the transient solution and it is very uninteresting to me. Assuming I start with x(0)=0 and v(0)=0 and a lightly damped system, then the highest magnitude reaches is at steady state. Changing the angle of application has little overall effect. As I increase the damping the steady state level decreases and also the time to reach the steady state level also decreases. I was a little surprised at this since I expected damping to &quot;slow down&quot; the time to reach steady state... but not so.

If you are only interested in the maximum value and the system meets conditions above (lightly damped and initial conditions are zero), then all you need to study is the steady state solution. By the way the steady state solution magnitude matched that predicted in my first post.

I'm still working on getting my transient solution into word.

 

[GregLocock]

Umm, I'll see your Laplace transform and raise. It should overshoot in the first couple of cycles, Shirley?

My Excel simulation does anyway (tho I'm not happy with it).
Cheers

Greg Locock

 

[electricpete]

Greg - you may be right but my simulation didn't show any overshoot whatsoever. I'm guessing that it takes time to put energy into the system.
One set of parameters I used (see attached) were k=1,m=1,c=0.1,theta=0.3. By the way with Fmax=1 this gave max displacement of 10 as predicted above. I tried several others that I thought were representative and saw similar behavior. But certainly could be my model is wrong or my parameter selection not good.

Here's my file:

http://geocities.com/pschimpf/mass_spring_transient_link.htm

If you are interested to email me your excel file then go the webpage listed on my profile and then near the bottom link to my electricpete@.... email. (can't list it here at the risk of getting spam-bombed). Also if you like I can post your excel file for others.

I agree it's fun. It is too few times during real work that I get to use those types of tools.

 

[electricpete]

Greg
I've been trying to think of physical examples which might shed some light on whether overshoot is expected for lightly damped system with resonant sinusoidal excitation applied suddenly.

I can think of one that seems somewhat close. If you hold one end of a heavy flexible yardstick you can get it to bend with only a slight movement of your wrist... at the right frequency. But when you first start even at the precise frequency there is no bending until you have given the amplitude a chance to build up. Not sure if that's exactly right (I think my forcing function is displacement at my end of the stick, not force at my end) but it seems close.

The electrical circuits I am familiar with that generate an overshoot are first-order systems. For instance a voltage applied to a series inductor/resistor circuit can cause current to go twice as high as the steady state value due to the decarying dc offset transient compoent.

An 2nd-order LC circuit generating an overshoot? No examples pop into my head offhand (but that doesn't mean there aren't any).

Can you think of any examples?

I for one have a hard time imagining an every-day life examples where resonance occurs and the forcing function is &quot;force&quot;. Part of the problem is it's tough for me to imagine applying force to something that's moving.

If you are pushing a child on a swing (a pendulum but similar to a mass spring I think), what is the best part of the swing cycle to push forward? Intuition tells me that it's when the child is all the way back. But if I apply resonance theory I would push when the swing is directly vertical (at it's lowest point in travel). That would equate to a forward peak in force occuring at a point in time 90 degrees ahead of the forward peak in travel. (force leads displacement by 90 degrees in resonance). But it's darned tough to push at the bottom because the kid is moving so fast. Intuition fails me.

Any comments.